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if $X$ and $Y$ are independent normal distribuited random variables and $T=2X-Y-1$ and $E[X]=E[Y]=1$ and $Var(X)=Var(Y)=4$, what is $Var(T)$? I get $E[T]=E[2X-Y-1]=2-1-1=0$, but i don't know how to get $Var(T)$.

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2 Answers 2

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This answer (which is more of an outline to an approach, cq. a hint) intends to indicate a means of deriving some useful properties about the variance $\operatorname{Var}(X)$.

It is not necessary to know that $X$ and $Y$ are normal, as long as we have the information that $\operatorname{Var}(X)$ and $\operatorname{Var}(Y)$ exist (and that $X,Y$ are independent).

Recall the definition of $\operatorname{Var}(X)$:

$$\operatorname{Var}(X) = \Bbb E[(X-\Bbb E X)^2] = \Bbb E[X^2] - \Bbb E[X]^2$$

where the last equality was derived using the following properties of $\Bbb E$, for independent $X, Y$ and $\lambda \in \Bbb R$:

\begin{align*} \Bbb E[X+\lambda Y] &= \Bbb E[X]+\lambda\Bbb E[Y]\\ \Bbb E[XY] &= \Bbb E[X]\Bbb E[Y] \end{align*}

(Note that $X$ is not independent of $X$ itself, so the last equation does not help for computing $\Bbb E[X^2]$!)

Using only these, it is a good and enlightening exercise to prove the following about the variance:

$$ \operatorname{Var}(X+\lambda Y) = \operatorname{Var}(X) + \lambda^2\operatorname{Var}(Y) $$


Once you've done this, computing $\operatorname{Var}(T)$ should be a piece of cake.

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  • $\begingroup$ I get : $Var(X)+\lambda^2Var(Y)-2 \lambda E[X]E[Y]$ $\endgroup$
    – blob
    Commented Jun 7, 2013 at 13:18
  • $\begingroup$ That looks like you've made a small mistake in computing $\Bbb E[(X+\lambda Y)^2] = \Bbb E[X^2+2\lambda XY+\lambda^2Y^2]$. $\endgroup$
    – Lord_Farin
    Commented Jun 7, 2013 at 13:20
  • $\begingroup$ yes i forgot $2 \lambda E[X]E[Y] - 2 \lambda E[X]E[Y]$, thank you. $\endgroup$
    – blob
    Commented Jun 7, 2013 at 13:25
  • $\begingroup$ I don't like Statistics but here, it smells good. :-) +1 $\endgroup$
    – Mikasa
    Commented Jun 8, 2013 at 10:54
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Use $$\operatorname{var}(aX+bY+c) = \operatorname{var}(aX+bY) = a^2\cdot\operatorname{var}(X)+b^2\cdot\operatorname{var}(Y)+2\cdot a \cdot b\cdot\operatorname{cov}(X,Y)$$ and the fact that $\operatorname{cov}(X,Y)=0$ for independent random variables $X$ and $Y$.

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  • $\begingroup$ thank you!, I've not yet read about the covariance. $\endgroup$
    – blob
    Commented Jun 7, 2013 at 13:03

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